Finding angles determined by given degrees

Question

I have more problems that I can not find the answer to. I have an answer key, so the issue is not the the answer but the explanation that is missing.

Thank you so much!

Answer 1

You can fill in any remaining details.

1. $AC$ is diameter. So $\angle ABC = 90^{\circ}$

Chord BC subtends equal angles $\angle BDC = \angle BAC$ on same side of circle.

$$ \alpha = \angle CAB = 90^{\circ} - 61^{\circ} = 29^{\circ} $$

2. In isosceles $\triangle AMB$, $\angle MAB = 48^{\circ} = \angle MBA$.

So $\angle AMB = 84^{\circ} = \angle MBC$.

$\angle ACB= \frac{1}{2}\angle AMB = 42^{\circ}$

$$\therefore \gamma = 180^{\circ} - 84^{\circ} - 42^{\circ}= 54^{\circ}$$

3. $ADCB$ is semicircle. So $\angle ADB = \angle ACB = 90^{\circ}$.

In isosceles $\triangle PMB$, $\angle PMB = \angle PBM = 74^{\circ}$

$$ \delta = \angle BAD - \angle BAC$$ $$ = \frac{1}{2}\angle BMD - (90^{\circ} - \angle ABC)$$ $$ = \frac{1}{2}\cdot74^{\circ} - (90^{\circ} - 74^{\circ})$$ $$ = 21^{\circ}$$